I came across these inequalities while learning about Schwartz functions (Classical Fourier Analysis, Grafakos) and I have no idea how to prove this:

For $x \in \mathbb{R}^{n}$ and $\alpha = (\alpha_{1}, \ldots, \alpha_{n}) \in \mathbb{N}^{n}$, we set

$$ x^{\alpha} = x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}.$$

Then prove that there exists a constant $c_{n,\alpha}$ such that

$$\left| x^{\alpha}\right| \leq c_{n,\alpha}|x|^{|\alpha|}$$

where $|\alpha| = \alpha_{1} + \cdots + \alpha_{n}$.

Conversely, for every $k \in \mathbb{N}$, there exists a $C_{n,k}$ such that

$$|x|^{k} \leq C_{n,k}\sum\limits_{|\beta| = k}|x^{\beta}|$$

Any help would be appreciated.


The first inequality is clear at $x=0$ with any constant . Now note that it suffices to prove it for $|x|=1$. This is because if it is true on the unit sphere then since $x\neq0$ then $y=(\frac{x_{1}}{|x|},\frac{x_{2}}{|x|},...,\frac{x_{n}}{|x|})=\frac{x}{|x|}\in S^{n-1}$ and then we have

$$\begin{align} \frac{|x^{\alpha}|}{|x|^{|\alpha|}} & = \frac{ \sqrt{ x_1^{2\alpha_1} + \dots + x_n^{2\alpha_n} } }{|x|^{|\alpha|} } \\ & = \bigg( \frac{ x_1^{2\alpha_1} + \dots + x_n^{2\alpha_n} }{|x|^{2|\alpha|}} \bigg)^{\frac{1}{2}} \\ & = \bigg( \frac{x_1^{2\alpha_1}}{|x|^{2|\alpha|}} + \dots + \frac{x_n^{2\alpha_n}}{|x|^{2|\alpha|}} \bigg)^{\frac{1}{2}} \\ & \leq \bigg[ \bigg(\frac{x_1}{|x|}\bigg)^{2\alpha_1} + \dots + \bigg(\frac{x_n}{|x|}\bigg)^{2\alpha_n}\bigg]^{\frac{1}{2}} \\ & = \bigg| \bigg( \frac{x_1^{\alpha_1}}{|x|^{\alpha_1}}, \dots, \frac{x_n^{\alpha_n}}{|x|^{\alpha_n}} \bigg) \bigg| \\ & = | y^{\alpha} | \\ & \leq c_{n,\alpha} |y|^{|\alpha|} \\ & = c_{n,\alpha}\bigg|\bigg(\frac{x_{1}}{|x|},\frac{x_{2}}{|x|},...\frac{x_{n}}{|x|}\bigg)\bigg|^{|\alpha|} \\ & = c_{n,\alpha}\frac{1}{|x|^{|\alpha|}}|(x_{1},x_{2},...x_{n})|^{|\alpha|} \\ & = c_{n,\alpha}\frac{|x|^{|\alpha|}}{|x|^{|\alpha|}} = c_{n,\alpha} \end{align}$$

So $|x^{\alpha}|\le c_{n,\alpha}|x|^{|\alpha|}$.

Now we prove this inequality on the unit sphere.

$\vert x^{\alpha}\vert=|x_1^{\alpha_1}||x_2^{\alpha_{2}}|...|x_n^{\alpha_n}|\le\frac{1}{n}\sum_{k=1}^{n}|x_{k}|^{n\alpha_{k}}\le\frac{1}{n}\big(\sum_{k=1}^{n}|x_{k}|^{\alpha_{k}}\big)^{n}$


Note that I have used the Arithmetic Geometric Inequality to get the first inequality in the above proof. The second inequality follows from the inequality: $(x_{1}+x_{2}+...+x_{n})^{k}=\sum_{\alpha_{1}+...+\alpha_{n}=k}\binom{k}{\alpha_{1}, \alpha_{2},...,\alpha_{n}}x^{\alpha_{1}}...x^{\alpha_{n}}\ge x_{1}^{k}+...+x_{n}^{k}$ which is true when $x_{1},...x_{n}$ are all non-negative. Also note that $|x_{k}|\le1+|x_{k}|$ and $|x_{k}|\le \sqrt{x_{1}^{2}+...+x_{n}^{2}}=|x|$ for all $k=1,2,...,n$.

For the second inequality we attack as follows:



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